Date modified: Thu 2023-06-22 . 06:12 PM
Date created: Sat 2023-10-07 . 09:02 PM
defining exp function calculus of exp hyperbolic functions l'hospital rule ## General log and general exponentiation with different bases. (6.4 $\ast$) restriction of domain; inverse trig and inverse hyperbolic ^ so we can integrate more functions deferred ## Logarithmic derivative. Often it is useful to compute the derivative of the natural logarithm of a function $f$ to find its derivative. We note that $\frac{d}{dx}\ln(f(x))= \frac{f'(x)}{f(x)}$. This is called the **logarithmic derivative** of $f$. This is useful whenever you want $f'$ but $f$ is a product of many things. Example. Compute $f'(x)$ if $$ f(x) = \frac{(3x^2+x-1)\sqrt{x^5+8}}{(x^3+2)(x+1)^{1 / 3}} $$ Ok. First take its natural log: $$ \ln(f(x) = \ln(3x^2+x-1)+\frac{1}{2}\ln(x^5 + 8)-\ln(x^3 + 2)- \frac{1}{3} \ln(x+1) $$ Then take the derivative of this log: $$ \frac{d}{dx}\ln f(x) = \frac{6x + 1}{3x^2 + x - 1} + \frac{5x^4}{2(x^5 + 8)} - \frac{3x^2}{x^3 + 2} - \frac{1}{3(x+1)} $$ So by the logarithmic derivative rule, $\frac{d}{dx}\ln(f(x))= \frac{f'(x)}{f(x)}$, we have $$ \begin{align*} &f'(x) = f(x) \cdot \frac{d}{dx}\ln(f(x))\\ &=\frac{(3x^2+x-1)\sqrt{x^5+8}}{(x^3+2)(x+1)^{1 / 3}}\left(\frac{6x + 1}{3x^2 + x - 1} + \frac{5x^4}{2(x^5 + 8)} - \frac{3x^2}{x^3 + 2} - \frac{1}{3(x+1)}\right) \end{align*} $$